To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution. By definition is defined as:
![\theta = 1.22\frac{\lambda}{d}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%201.22%5Cfrac%7B%5Clambda%7D%7Bd%7D)
Where,
= Wavelength
d = Width of the slit
= Angular resolution
Through the arc length we can find the radius, which would be given according to the length and angle previously described.
The radius of the beam on the moon is
![r = l\theta](https://tex.z-dn.net/?f=r%20%3D%20l%5Ctheta)
Relacing ![\theta](https://tex.z-dn.net/?f=%5Ctheta)
![r = l(\frac{1.22\lambda}{d})](https://tex.z-dn.net/?f=r%20%3D%20l%28%5Cfrac%7B1.22%5Clambda%7D%7Bd%7D%29)
![r = 1.22\frac{l\lambda}{d}](https://tex.z-dn.net/?f=r%20%3D%201.22%5Cfrac%7Bl%5Clambda%7D%7Bd%7D)
Replacing with our values we have that,
![r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})](https://tex.z-dn.net/?f=r%20%3D%201.22%2A%28%5Cfrac%7B%28384%2A10%5E3km%29%28%5Cfrac%7B1000m%7D%7B1km%7D%29%28550%2A10%5E%7B-9%7Dm%29%7D%7B7%2A10%5E%7B%7B-2%7D%7D%7D%29)
![r = 3680.91m](https://tex.z-dn.net/?f=r%20%3D%203680.91m)
Therefore the diameter of the beam on the moon is
![d = 2r](https://tex.z-dn.net/?f=d%20%3D%202r)
![d = 2 * (3690.91)](https://tex.z-dn.net/?f=d%20%3D%202%20%2A%20%283690.91%29)
![d = 7361.8285m](https://tex.z-dn.net/?f=d%20%3D%207361.8285m)
Hence, the diameter of the beam when it reaches the moon is 7361.82m
Answer:
F = 8 N
Explanation:
The question says "The body is subjected to a force with a moment of 0.4 N × m shoulder - 5 cm. What is the magnitude of this force?
Given that,
Moment/Torque, ![\tau=0.4\ N-m](https://tex.z-dn.net/?f=%5Ctau%3D0.4%5C%20N-m)
Distance moved, d = 5 cm = 0.05 m
We need to find the magnitude of this force. We know that, the torque acting on an object is given by :
![\tau=Fd\\\\F=\dfrac{\tau}{d}\\\\F=\dfrac{0.4\ N-m}{0.05\ m}\\\\F=8\ N](https://tex.z-dn.net/?f=%5Ctau%3DFd%5C%5C%5C%5CF%3D%5Cdfrac%7B%5Ctau%7D%7Bd%7D%5C%5C%5C%5CF%3D%5Cdfrac%7B0.4%5C%20N-m%7D%7B0.05%5C%20m%7D%5C%5C%5C%5CF%3D8%5C%20N)
So, the magnitude of force is equal to 8 N.
Compounds are molecules with 2 or more elements
So the answer would be the third one
CO2;H2O
Answer:
m₂ = 1.33 kg
Explanation:
1.50 rev/s = 3π rad/s
I₁ = ½(2.00)(0.50²) = 0.25 kg•m²
L₁ = I₁ω₁ = 0.25(3π) = .075π kg•m²/s
ω₂ = ω₁R₁/R₂ = 3π(0.5/0.75) = 2π rad/s
I₂ = L₂/ω₂ = 0.75π/2π = 0.375 kg•m²
I₂ = ½m₂R₂²
m₂ = 2I₂/R₂² = 2(0.375) / 0.75² = 1.33 kg